Ellipse Inscribed in a Trapezoid

Utilizing symmetry, we can consider only one half of the given figure,

We apply horizontal scaling to bring half the ellipse into a semi-circle.

The scaled upper base will have a length of $(4s)$ , and the lower base will have a length of $(10s)$ . where $s$ is the scale factor.

Using the result of this problem , we have the following relation,

$(4s)(10s) = (\text{radius of semi-circle})^2 = 4^2 = 16$

Hence $s = \sqrt{ \dfrac{2}{5} }$

Hence ellipse semi-major axis $= a = \dfrac{4}{s} = 4 \sqrt{ \dfrac{5}{2} } = \sqrt{40}$

From which, the major axis $= 2 a = 2 \sqrt{40} = \boxed{12.649}.$